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Figure 2 shows a sketch of part of the curve C with equation y = 2ln(2x + 5) - \frac{3x}{2} \quad , \quad x > -2.5 The point P with x coordinate -2 lies on C - Edexcel - A-Level Maths Pure - Question 5 - 2017 - Paper 4
Question 5
Figure 2 shows a sketch of part of the curve C with equation y = 2ln(2x + 5) - \frac{3x}{2} \quad , \quad x > -2.5 The point P with x coordinate -2 lies on C. (a)... show full transcript
Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve C with equation y = 2ln(2x + 5) - \frac{3x}{2} \quad , \quad x > -2.5 The point P with x coordinate -2 lies on C - Edexcel - A-Level Maths Pure - Question 5 - 2017 - Paper 4
Step 1
Find an equation of the normal to C at P
Answer
First, we determine the y-coordinate of point P when x = -2:
Thus, the coordinates of P are (-2, 3).
Next, we find the derivative of the curve to calculate the slope of the tangent line at point P:
[ \frac{dy}{dx} = \frac{4}{2x+5} - \frac{3}{2} \quad \text{at } x = -2 \text{ gives } \frac{dy}{dx} = \frac{4}{1} - \frac{3}{2} = 4 - 1.5 = 2.5.]
The slope of the normal line is the negative reciprocal of the tangent slope:
[ m_{normal} = -\frac{1}{2.5} = -\frac{2}{5}. ]
Using point-slope form to find the equation of the normal line:
[ y - 3 = -\frac{2}{5}(x + 2). ]
Rearranging gives:
[ 2x + 5y = 11, ] where a = 2, b = 5, c = 11.
Step 2
Show that the x coordinate of Q is a solution of the equation
Answer
Using the normal line equation, we express the curve's equation:
[ 5y + 2x = 11 ext{ and } y = 2ln(2x + 5) - \frac{3x}{2}. ]
Substituting the expression for y:
[ 5(2ln(2x + 5) - \frac{3x}{2}) + 2x = 11,] which simplifies to [ 10ln(2x + 5) - \frac{15x}{2} + 2x = 11. ]
Combining terms leads to: [ 10ln(2x + 5) - \frac{11x}{2} = 11. ]
Rearranging the equation gives: [ x = \frac{20}{11}ln(2x + 5) - 2.]
Step 3
Taking x_1 = 2, find the values of x_n and x_{n+1}
Answer
Starting with x_1 = 2, we calculate x_2 using the iteration formula:
[ x_{n+1} = \frac{20}{11} ln(2x_{n} + 5) - 2. ]
Substituting x_1:
[ x_2 = \frac{20}{11} ln(2(2) + 5) - 2 = \frac{20}{11} ln(9) - 2. ]
Calculating:
- First, find ln(9) which is approximately 2.1972. Then, [ x_2 = \frac{20}{11}(2.1972) - 2 \approx 0.9953.]
Next, using x_2 to find x_3: [ x_3 = \frac{20}{11} ln(2(0.9953) + 5) - 2 = \frac{20}{11} ln(6.9906) - 2.]
This leads to:
- ln(6.9906) approximately 1.9313. [ x_3 = \frac{20}{11}(1.9313) - 2 \approx 0.7577.]
Continuing this process, we repeat the iterations to arrive at approximate values for x_n and x_{n+1}.